Answer
$70\pm5^{\circ}F$
Work Step by Step
Step 1. Given the equation $y=10+(t-70)\times 10^{-4}$, $y_0=10cm, t_0=70^{\circ}F$ and $\Delta y=0.0005cm$, we need to find out the range of temperatures to satisfy the conditions.
Step 2. The range of the length can be found as $y_1=y_0-\Delta y=10-0.0005=9.9995cm$ and $y_2=y_0+\Delta y=10+0.0005=10.0005cm$.
Step 3. Rewrite the equation as $t=70+(y-10)\times 10^4$.
For the given y-values, we can calculate the corresponding temperatures: $t_1=70+(y_1-10)\times 10^4=70+(9.9995-10)\times 10^4=65$ and $t_2=70+(y_2-10)\times 10^4=70+(10.0005-10)\times 10^4=75$
Step 4. Thus, the range of the temperatures needs to be $[60,75]^{\circ}F$ or $70\pm5^{\circ}F$
